Chemistry & Physics News

Tuesday, April 29, 2008

Rigid Rotation in 2d and 3d

After having studied translation and vibration in the particle-in-a-box and harmonic oscillator quantum systems, respectively, it is natural to turn to rotation. The simplest such system is the particle on a ring (also called rigid rotation in 2d) which, after replacing the mass m with the reduced massμ can be used for two-body problems (like a diatomic molecule).

Since the potential is a constant we can define it to be zero. The natural coordinate system to use is polar coordinates which greatly simplifies the Schrödinger Equation:

(-hbar2/2I) ∂2/∂φ2ψ = Eψ

This is identical to the very first problem we solved, expect we have traded translational properties (x, m) with rotational analogs (φ, I), a standard practice seen even in classical physics. This leads to, after normalizing, the following equation:

ψ(φ) = 1/√2π eimφ

Continuity conditions demands that cyclic boundary conditions hold; that is, that ψ(φ + 2π) = ψ(φ). This restriction leads directly to quantization, and the quantum number ml can only take integral values. This restriction on ml requires that both energy and position/displacement be quantized.

Further, after constructing the appropriate operator, we see that we have a new quantized property arises naturally in angular momentum, called Lzbecause the vector lies perpendicular to the plane of rotation.

Rotation in 3d is very similar but we need to use spherical coordinates...

4 comments:

Ian
said...

For the radial part of the wavefunction, we are given two different equations. On the equation sheet from the quiz, we have the Laguerre equation as a function of rho. However, on the handouts in class and notes, it is a function of rho/n. Which is it?

When you're referring to probability, the value can never be negative; thus it can only be zero or positive. For this reason, we often don't even bother writing the +'s and indicate the nodes with a line (just as in the wavefunction plots).